Assuming that the reactants are:
(NH4)2SO4 (aq) + Ba(NO3)2 (aq)
and the products are:
BaSO4 (s) + 2NH4NO3 (aq),
then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.
According to the solubility rules, the following elements are considered insoluble when paired with SO4:
Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+
Therefore, the precipitate will be BaSO4 (s).
1 mole of N2 produces 2 moles of NH3
OR...
14 x 2 grams of N2 produces 2(14 +3) grams of NH3
1 gram of N2 produces 34/28 grams of NH3
therefore, 56 grams produce (34/28 )x 56 =68 grams of NH3
the answer thus would be 68 grams of NH3
Answer:
Explanation:
Molarity = number of moles / volume
If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, which is half of 550 mL, the molarity of the solution with the same number of moles of KCl is 3.5 * 2 = 7.00 M
<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
therefore number of BaSO4 moles formed - 0.288 mol</span>
